Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(C\) to predict how \(\Delta G\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at $800 \mathrm{~K} ?\( (c) Calculate \)\Delta G\( at \)1000 \mathrm{~K}$. Is the reaction spontaneous under standard conditions at this temperature?

Short Answer

Expert verified
(a) The Gibbs Free Energy equation is given by \(\Delta G = \Delta H - T \Delta S\). (b) At 800 K, we have \(\Delta H = 55.8 \: kJ\), \(\Delta S = 173.5 \: J/K\), and \(\Delta G = -84,000 \: J/mol\). The reaction is spontaneous at 800 K, as \(\Delta G\) is negative. (c) At 1000 K, \(\Delta G = -117,700 \: J/mol\), which means the reaction is also spontaneous under standard conditions at 1000 K.

Step by step solution

01

a) Knowing the Gibbs Free Energy Equation:

We need to know the Gibbs Free Energy equation which is as follows: \[ \Delta G = \Delta H - T \Delta S \] where ∆G is the change in Gibbs free energy, ∆H is the change in enthalpy, T is the temperature in Kelvin, and ∆S is the change in entropy.
02

b) Calculating ∆G at 800 K:

We are given that ∆H° and ∆S° do not change with temperature. Using the data given in Appendix C, we have: ∆H°(NO2) = 33.1 kJ/mol, ∆S°(NO2) = 240.0 J/mol·K ∆H°(N2O) = 82.0 kJ/mol, ∆S°(N2O) = 218.6 J/mol·K ∆H°(NO) = 90.3 kJ/mol, ∆S°(NO) = 210.7 J/mol·K Now, we have to calculate the ∆H and ∆S of the reaction by using stoichiometric coefficients. ∆H = [3∆H°(NO)] - [∆H°(NO2) + ∆H°(N2O)] = (3*90.3) - (33.1+82.0) = 55.8 kJ ∆S = [3∆S°(NO)] - [∆S°(NO2) + ∆S°(N2O)] = (3*210.7) - (240.0+218.6) = 173.5 J/K Now we can calculate ∆G at 800 K: ∆G = ∆H - T∆S = 55.8*10^3 J/mol - (800 K)(173.5 J/mol·K) = -84,000 J/mol As ∆G is negative, the reaction is spontaneous at 800 K.
03

c) Calculating ∆G at 1000 K:

We already have ∆H and ∆S for the reaction; we only need to change the temperature in the equation for Gibbs Free Energy. ∆G = ∆H - T∆S = 55.8*10^3 J/mol - (1000 K)(173.5 J/mol·K) = -117,700 J/mol As ∆G is negative at 1000 K, the reaction is spontaneous under standard conditions at this temperature as well.

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Most popular questions from this chapter

(a) What do you expect for the sign of \(\Delta S\) in a chemical reaction in which 3 mol of gaseous reactants are converted to 2 mol of gaseous products? (b) For which of the processes in Exercise 19.11 does the entropy of the system increase?

Which of the following processes are spontaneous: (a) the evaporation of water at \(\$ T P\) to form water vapor of 101.3 kPa pressure; (b) separation of a mixture of water and oil into two separate phases; (c) the souring of milk; (d) the neutralization of hydrochloric acid with sodium hydroxide at \(\mathrm{STP} ;(\mathbf{e})\) the formation of ice from water at \(20^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa} ?\)

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Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) $\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)$ (b) $2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)$ (c) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$

Using data from Appendix \(\mathrm{C}\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at $298 \mathrm{~K}$ under standard conditions. (a) \(2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)\) (b) \(2 \mathrm{NaBr}(s) \longrightarrow 2 \mathrm{Na}(g)+\mathrm{Br}_{2}(g)\) (c) $\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)$

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